# Problem Restatement A drum filled with liquid (density \( \rho \)) to a height \( d \) has a hole at the center of its lid. The drum is rotated at angular velocity \( \omega \) (I assume \( \alpha \) in your question is a typo for \( \omega \)). We are to find the **inner radius** \( r_i \) (where the liquid contacts the lid—i.e., the *radius of the central "hole"* in the liquid surface that forms due to rotation). You provided the answer: \[ r_i = \left[ \frac{4 g r_o^2 (d_o - d)}{\omega^2} \right]^{1/4} \] where: - \( r_o \): radius of the drum - \( d_o \): initial height of the liquid (before rotation) - \( d \): height at the lowest point (center) after rotation - \( g \): acceleration due to gravity - \( \omega \): angular velocity Let's derive this step by step. --- # Step 1: Shape of Liquid Surface in Rotation When a liquid rotates at angular velocity \( \omega \), its free surface forms a paraboloid. The height of the liquid surface at any radius \( r \): \[ z(r) = z() + \frac{\omega^2}{2g} r^2 \] Where: - \( z() \): height at the center (\( r = \)) - \( z(r) \): height at radius \( r \) --- # Step 2: The "Hole" at the Center The drum lid has a central hole. As the drum spins, the liquid is pushed outward, and a central "hole" forms in the liquid surface, with inner radius \( r_i \). At \( r = r_i \), the liquid just contacts the lid (i.e., the surface touches the underside of the lid). Beyond \( r_i \), the liquid rises in a paraboloid up to the wall at \( r = r_o \). --- # Step 3: Conservation of Volume **The volume of the liquid does not change.** - Initial volume (before rotation): \[ V_{\text{initial}} = \pi r_o^2 d_o \] - Volume after rotation (paraboloid with central hole): \[ V_{\text{final}} = \text{Volume between } r_i \text{ and } r_o \] \[ V_{\text{final}} = \int_{r_i}^{r_o} 2\pi r \left[ z() + \frac{\omega^2}{2g} r^2 \right] dr \] But since the surface at \( r = r_i \) is at the height of the lid (i.e., \( z(r_i) = \)), set \( z() \) as \( -\frac{\omega^2}{2g} r_i^2 \): \[ z(r) = z() + \frac{\omega^2}{2g} r^2 = -\frac{\omega^2}{2g} r_i^2 + \frac{\omega^2}{2g} r^2 = \frac{\omega^2}{2g} (r^2 - r_i^2) \] --- # Step 4: Compute Final Volume \[ V = \int_{r_i}^{r_o} 2\pi r \cdot \frac{\omega^2}{2g} (r^2 - r_i^2) dr \] \[ = \frac{\pi \omega^2}{g} \int_{r_i}^{r_o} r (r^2 - r_i^2) dr \] \[ = \frac{\pi \omega^2}{g} \int_{r_i}^{r_o} (r^3 - r_i^2 r) dr \] \[ = \frac{\pi \omega^2}{g} \left[ \frac{r^4}{4} - \frac{r_i^2 r^2}{2} \right]_{r_i}^{r_o} \] \[ = \frac{\pi \omega^2}{g} \left\{ \left( \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} \right) - \left( \frac{r_i^4}{4} - \frac{r_i^4}{2} \right) \right\} \] \[ = \frac{\pi \omega^2}{g} \left[ \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} + \frac{r_i^4}{4} \right] \] --- # Step 5: Equate Initial and Final Volumes Set initial and final volumes equal: \[ \pi r_o^2 d_o = \frac{\pi \omega^2}{g} \left[ \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} + \frac{r_i^4}{4} \right] \] Divide both sides by \( \pi \): \[ r_o^2 d_o = \frac{\omega^2}{g} \left[ \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} + \frac{r_i^4}{4} \right] \] --- # Step 6: Solve for \( r_i \) Multiply both sides by \( \frac{g}{\omega^2} \): \[ r_o^2 d_o \frac{g}{\omega^2} = \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} + \frac{r_i^4}{4} \] Bring all terms to one side: \[ = \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} + \frac{r_i^4}{4} - r_o^2 d_o \frac{g}{\omega^2} \] Multiply both sides by 4: \[ = r_o^4 - 2 r_i^2 r_o^2 + r_i^4 - 4 r_o^2 d_o \frac{g}{\omega^2} \] Group terms: \[ r_i^4 - 2 r_i^2 r_o^2 + r_o^4 - 4 r_o^2 d_o \frac{g}{\omega^2} = \] This is a quadratic in \( r_i^2 \): \[ (r_i^2)^2 - 2 r_o^2 r_i^2 + r_o^4 - 4 r_o^2 d_o \frac{g}{\omega^2} = \] Let \( x = r_i^2 \): \[ x^2 - 2 r_o^2 x + r_o^4 - 4 r_o^2 d_o \frac{g}{\omega^2} = \] --- # Step 7: Solve the Quadratic The quadratic formula for \( x^2 + bx + c = \) is \( x = \frac{-b \pm \sqrt{b^2 - 4c}}{2} \): Here, \( b = -2 r_o^2 \), \( c = r_o^4 - 4 r_o^2 d_o \frac{g}{\omega^2} \): \[ x = \frac{2 r_o^2 \pm \sqrt{(2 r_o^2)^2 - 4 [r_o^4 - 4 r_o^2 d_o \frac{g}{\omega^2}]}}{2} \] Compute discriminant: \[ (2 r_o^2)^2 - 4 [r_o^4 - 4 r_o^2 d_o \frac{g}{\omega^2}] = 4 r_o^4 - 4 r_o^4 + 16 r_o^2 d_o \frac{g}{\omega^2} = 16 r_o^2 d_o \frac{g}{\omega^2} \] Thus, \[ x = \frac{2 r_o^2 \pm \sqrt{16 r_o^2 d_o \frac{g}{\omega^2}}}{2} \] \[ x = \frac{2 r_o^2 \pm 4 \sqrt{r_o^2 d_o \frac{g}{\omega^2}}}{2} \] \[ x = r_o^2 \pm 2 \sqrt{r_o^2 d_o \frac{g}{\omega^2}} \] We take the **smaller** root, since the "hole" is at the center: \[ r_i^2 = r_o^2 - 2 \sqrt{r_o^2 d_o \frac{g}{\omega^2}} \] But this does **not** match your formula. Let's check for a possible error. --- # Step 8: Use Difference in Heights Alternatively, we can relate the difference in heights at the edge and at the center. From the paraboloid profile: \[ z(r) = z() + \frac{\omega^2}{2g} r^2 \] At \( r_o \): \( z(r_o) \) At \( r_i \): \( z(r_i) = \) (contacts lid) So difference in height between edge and center: \[ z(r_o) - z(r_i) = \frac{\omega^2}{2g} (r_o^2 - r_i^2) \] But the liquid at the edge rises by \( d_o - d \): \[ d_o - d = \frac{\omega^2}{2g} (r_o^2 - r_i^2) \] So, \[ r_o^2 - r_i^2 = \frac{2g}{\omega^2} (d_o - d) \] \[ r_i^2 = r_o^2 - \frac{2g}{\omega^2} (d_o - d) \] Now, the **volume** of the liquid must equal the original volume: \[ \text{Initial: } \pi r_o^2 d_o \] \[ \text{Final: } \int_{r_i}^{r_o} 2\pi r [ z(r)] dr \] But since \( z(r_i) = \): \[ z(r) = \frac{\omega^2}{2g} (r^2 - r_i^2) \] So, \[ V = \int_{r_i}^{r_o} 2\pi r \cdot \frac{\omega^2}{2g} (r^2 - r_i^2) dr \] \[ = \frac{\pi \omega^2}{g} \int_{r_i}^{r_o} r(r^2 - r_i^2) dr \] \[ = \frac{\pi \omega^2}{g} \left[ \frac{r^4}{4} - \frac{r_i^2 r^2}{2} \right]_{r_i}^{r_o} \] As before, \[ V = \frac{\pi \omega^2}{g} \left[ \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} - \left( \frac{r_i^4}{4} - \frac{r_i^4}{2} \right) \right] \] \[ = \frac{\pi \omega^2}{g} \left[ \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} + \frac{r_i^4}{4} \right] \] Set equal to original volume: \[ \pi r_o^2 d_o = \frac{\pi \omega^2}{g} \left[ \frac{r_o^4}{4} - \frac{r_i^2 r_o^2}{2} + \frac{r_i^4}{4} \right] \] So as above, but now substitute \( r_i^2 = r_o^2 - \frac{2g}{\omega^2}(d_o - d) \): Let’s denote \( \Delta d = d_o - d \). Let’s substitute \( r_i^2 \) into the volume equation: \[ r_o^2 d_o = \frac{\omega^2}{g} \left[ \frac{r_o^4}{4} - \frac{[r_o^2 - \frac{2g}{\omega^2}\Delta d] r_o^2}{2} + \frac{[r_o^2 - \frac{2g}{\omega^2}\Delta d]^2}{4} \right] \] Let’s define \( A = r_o^2 \), \( B = \frac{2g}{\omega^2}\Delta d \): \[ r_i^2 = A - B \] Now, plug back into the equation: \[ A d_o = \frac{\omega^2}{g} \left[ \frac{A^2}{4} - \frac{(A - B)A}{2} + \frac{(A - B)^2}{4} \right] \] Expand: \[ \frac{A^2}{4} - \frac{A^2 - AB}{2} + \